weak base ionization help

Question

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.66

A. Determine the concentration of C6H5NH3+ in the solution if the concentration of C6H5NH2 is 0.305 M. The p𝐾bp�b of aniline is 9.13.

B. Calculate the change in pH of the solution, ΔpH, if 0.360 g NaOH is added to the buffer for a final volume of 1.15 L. Assume that any contribution of NaOH to the volume is negligible.

Answer 1

A. Henderson Hasselbalch equation: pH = pKa + log [conj.base] / [acid]

C₆H₅NH₂ = base

C₆H₅NH₃⁺ = acid

pH = 5.66

pKa = 14 - pKb = 14 - 9.13 = 4.87

Solve for [conj.base]:

5.66 = 4.87 + log [C₆H₅NH₂] / [C₆H₅NH₃⁺]

0.79 = log [C₆H₅NH₂] / [C₆H₅NH₃⁺]

[C₆H₅NH₂] / [C₆H₅NH₃⁺] = 6.17

0.305 / [C₆H₅NH₃⁺] = 6.17

[C₆H₅NH₃⁺] = 0.0494 M

B. Concentration of NaOH added = 0.360 g NaOH x 1 mol / 40 g / 1.15 L = 0.00782 M

NaOH will react with C₆H₅NH₃⁺ to produce C₆H₅NH₂

Final [C₆H₅NH₃⁺] = 0.0494 M - 0.00782 M = 0.04158 M

Final [C₆H₅NH₂] = 0.305 M + 0.00782 M = 0.3128 M

pH = pKa + log [C₆H₅NH₂] / [C₆H₅NH₃⁺]

pH = 4.87 + log (0.3128 / 0.04158) = 4.87 + 0.88

pH = 5.75

∆pH = 0.88 (or 5.75 - 4.87)